# KPK 9th Class Physics Chapter 6 Work and Energy Short Questions Answers

KPK 9th Class Physics Chapter 6 Work and Energy Short Questions with answers are combined for all 9th class(Matric/ssc) Level students.Here You can prepare all Physics Chapter 6 Work and Energy short question in unique way and also attempt quiz related to this chapter.Just Click on Short Question and below Answer automatically shown. After each question you can give like/dislike to tell other students how its useful for each.

**Class/Subject: 9th Class ****Physics**

**Chapter Name: Work and Energy**

**Board: All ****KPK Boards**

- Malakand Board 9th Class Physics Chapter 6 Work and Energy short questions Answer
- Mardan Board 9th Class Physics Chapter 6 Work and Energy short questions Answer
- Peshawar Board 9th Class Physics Chapter 6 Work and Energy short questions Answer
- Swat Board 9th Class Physics Chapter 6 Work and Energy short questions Answer
- Dera Ismail Khan Board 9th Class Physics Chapter 6 Work and Energy short questions Answer
- Kohat Board 9th Class Physics Chapter 6 Work and Energy short questions Answer
- Abbottabad Board 9th Class Physics Chapter 6 Work and Energy short questions Answer
- Bannu Board 9th Class Physics Chapter 6 Work and Energy short questions Answer

**Helpful For:**

- All KPK Boards 9th Class Physics Annual Examination
- Schools 9th Class Physics December Test
- KPK 9th Class Physics Test
- Entry Test questions related Physics

## KPK 9th Class Physics Chapter 6 Work and Energy Short Questions Answers

Can a centripetal force ever do work on an object? Explain.

No, centripetal force never do any work on an object. The centripetal force compels a body to move in a circle. The centripetal force is always perpendicular to the direction of the motion. i.e 0 = 90°

And we know that:

Work = FS cos0

= FScos90° cos90°

Work = 0

To do any work the components of the force must be in the direction of motion. The centripetal force has no such components, so centripetal force never do work.

What happens to the kinetic energy of a bullet when it penetrates into a sand bag?

When a bullet is penetrating into a sand bag, sand opposes the motion of the bullet as a retarding force. So the kinetic energy of the bullet is utilized against the retarding force of the sand.

A meteor enters into earth atmosphere and burns, what happens to its kinetic energy?

When a meteor enters into earth atmosphere and burns kinetic energy of meteor is then converted into Heat and Light energy.

Two bullets are fired at the same time with same kinetic energy. If one bullet has twice the mass of the other, which has the greater speed and by what factor? Which can do the most work?

Consider two bullets having masses m1 and m2 and moving with velocities v

(K.E)

Similarly the K.E of bullet of mass ‘m2‘ will be?”:

(K.E)

Since both the bullets have same K.E but the mass of one bullet is twice of the other bullet (i.e m

Now,

(K.E)

2m

√v

The bullet of mass “m

_{1}and v_{2 }K.E of the bullet of mass m1 will be:(K.E)

_{1}=1/2m_{1}v_{1}^{2}…………(1)Similarly the K.E of bullet of mass ‘m2‘ will be?”:

(K.E)

_{2}=1/2m_{2}v_{2}^{2}………… (2)Since both the bullets have same K.E but the mass of one bullet is twice of the other bullet (i.e m

_{1}=2m_{2}).Now,

(K.E)

_{1}= (K.E)_{2 }1/2m_{1}v_{1}^{2}=1/2m_{2}v_{2}^{2 }1/2(2m_{2})v_{1}2=1/2m_{2}v_{2}^{2}(∴m_{1}=2m_{2})2m

_{2}v_{1}^{2}=m_{2}v_{2}^{2 }2m_{2}v_{1}^{2}/m^{2}= v_{2}^{2 }v_{2}^{2}= 2v_{1}^{2 }Talking square root on both sides:√v

_{2}^{2}= √2√v_{1}^{2 }v_{2}= √2v_{1 }Thus the bullet of mass “m_{2}“ which is lighter than “m1” travels with great speed with a facts of √2.The bullet of mass “m

_{2}” covers a greater distance than “m_{1}” so bullet having mass “m_{2}” do the most work.Can an object have different amounts of gravitational potential energy if it remains at the same elevation?

No, an object cannot have different amount of gravitational potential energy. If it remains at the same elevation we know that

P.E= mgh

P.E depends upon the height from the surface of earth. If the object remains at the same elevational from the surface of earth the gravitational P.E will remain same. Gravitational P.E is increased with the increase of height.

P.E= mgh

P.E depends upon the height from the surface of earth. If the object remains at the same elevational from the surface of earth the gravitational P.E will remain same. Gravitational P.E is increased with the increase of height.

Why do road leading to the top of a mountain wind back and forth?

1. When the road are wind up, due to the small angle of inclination we can do less work against the gravity instead of straight up situation.

2. By increasing the frictional force between road and tyres the road leading to the top of a mountain wind back and forth.

3. To avoid the skidding force the road leading to top a mountain wind back and forth.

Which would have a greater effect on the K.E of an object, doubling the mass or doubling the velocity?

Since

K.E=1/2 mv

^{2 }If we double the mass only then m’ =2m we get.

K.E‘=1/2×2m ×v

^{2 }K.E=2 (1/2mv

^{2})

(K.E) = 2K.E

If we double the velocity only then:

V “ = 2v

(K.E)” = 1/2 m (2v)

^{2 }(K.E)” = 1/2×m×4v

^{2 }(K.E)” = 4(1/2mv

^{2})

(K.E)” = 4 K.E

This result shows that greater effect on the kinetic energy is produced by doubling the velocity.

If the speed of a particle triples, by what factor does its kinetic energy increases?

Kinetic energy (K.E) of the particle of mass “m is given by:

K.E = 1/2mv

If the speed of the particle is tripled then V’ = 3V.

Now equation (1) can be written as:

(K.E)’ =1/2mv

K.E)’ = 1/2m (3v)

(K.E)’ = 9K.E (∴K.E=1/2mv

When the speed of the particle increase to triple then K.E increases 9 times.

K.E = 1/2mv

^{2}……………(1)If the speed of the particle is tripled then V’ = 3V.

Now equation (1) can be written as:

(K.E)’ =1/2mv

^{2 }Put v’ = 3vK.E)’ = 1/2m (3v)

^{2 }= 1/2m9.v^{2 }= 9(1/2mmv^{2})(K.E)’ = 9K.E (∴K.E=1/2mv

^{2})When the speed of the particle increase to triple then K.E increases 9 times.

The motor of a carne use power P to lift a steel beam. By what factor must the motor’s power increase to lift the beam twice as high in halt the time?

The rate of doing work is called power.

Power = work/time

Here work = P.E =mgh

Power = mgh/t…………..(1)

When height of beam because double in half time, then above equation can be written as:

P’ = mgh‘/t‘ …………………(2)

Where h’ = 2h

And t’=t/2

Now equ: (2) becomes:

P’ = mg×2h/t

P’ = 2mg×2/t

P’ = 4mgh/t

P’=4P

∴P= mgh/t

The motor’s power increases by four times.